JEE Main 2022 · 28 Jun · Shift 1 · Mathematics Q19

← JEE Main 2022 · 28 Jun · Shift 1

MathematicsSingle correct+4 / -1Question 19 of 90

Let AB and PQ be two vertical poles, 160 m apart from each other. Let C be the middle point of B and Q, which are feet of these two poles. Let $\dfrac\pi8$ and $\theta$ be the angles of elevation from C to P and A, respectively. If the height of pole PQ is twice the height of pole AB, then $\tan^2\theta$ is equal to

\dfrac{3-2\sqrt2}2

\dfrac{3+\sqrt2}2

\dfrac{3-2\sqrt2}4

\dfrac{3-\sqrt2}4

Answer and worked solution

Official answer: C

Heights. C is 80 m from each foot. $PQ=80\tan\frac\pi8=80(\sqrt2-1)$ , so $AB=40(\sqrt2-1)$ .

Angle. $\tan\theta=\dfrac{AB}{80}=\dfrac{\sqrt2-1}2$ , so $\tan^2\theta=\dfrac{3-2\sqrt2}4$ .

(C)

Topics:Heights and distances Trigonometric ratios and identities

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JEE Main 2022 · 28 Jun · Shift 1, Mathematics question 19